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Suppose $M$ and $N$ are free $R$-module($R$ is a commutative ring). The tensor product of $M\otimes_R N$ is free $R$-module? I know for projective modules it is true. How should we build its basis?

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    $\begingroup$ A basis is the most natural possible: the tensor product of bases. $\endgroup$ – user26857 May 14 '17 at 19:11
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$$[\bigoplus\limits_{i} R] \otimes_R [ \bigoplus\limits_j R] \cong \bigoplus\limits_{i,j} R \otimes_R R \cong \bigoplus\limits_{i,j} R$$

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  • $\begingroup$ That's right. And what about basis? $\endgroup$ – B.K-Theory May 14 '17 at 19:15
  • $\begingroup$ If you go through those isomorphisms explicitly, you should be able to see what the basis is $\endgroup$ – D_S May 14 '17 at 19:17
  • $\begingroup$ Could we say its basis come from the tensor product of basis of $M$ and $N$? $\endgroup$ – B.K-Theory May 14 '17 at 19:21

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